Page 168 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 10] STOICHIOMETRY 157

◦
◦
◦
10.43. How much heat is required to raise 43.2 g of iron from 20.0 Cto35.1 C? [c = 0.447 J/(g· C)]
Ans. t = 35.1 C − 20.0 C = 15.1 C
◦
◦
◦

0.447 J
Heat = (m)(c)( t) = (43.2g) (15.1 C) = 292 J
◦
g· C
◦
10.44. Calculate the heat produced by incomplete combustion of carbon, producing 45.7 g of CO according to
the following equation:
2C + O 2 −→ 2CO + 220 kJ

Ans. 45.7gCO 1 mol CO 220 kJ = 180 kJ
28.0gCO 2 mol CO
10.45. Calculate the temperature change produced by the addition of 225 J of heat to 15.3 g of water.
225 J
Ans. t = = 3.51 C
◦
[4.184 J/(g· C)](15.3g)
◦
10.46. Calculate the final temperature t f if 225 J of energy is added to 15.3 g of water at 19.0 C.
◦
Ans. t was calculated in the prior problem.
◦
◦
◦
t f = 19.0 C + 3.51 C = 22.5 C
10.47. Calculate the specific heat of a 125 g metal bar which rose in temperature from 18.0 C to 33.0 Con
◦
◦
addition of 2.37 kJ of heat.
heat 2370 J
Ans. c = = = 1.26 J/(g· C)
◦
(m)(t) (125 g)(15.0 C)
◦
Supplementary Problems

10.48. Read each of the other supplementary problems and state which ones are limiting-quantities problems.
Ans. 10.57, 10.58, 10.59, 10.64, 10.65, 10.66, 10.67, 10.71.

10.49. Calculate the number of grams of methyl alcohol, CH 3 OH, obtainable in an industrial process from 5.00 metric tons
6
(5.00 × 10 g) of CO plus hydrogen gas.
Ans. See Problem 10.21.

10.50. Make up your own stoichiometry problem, using the equation
CaCl 2 + 2 AgNO −→ Ca(NO 3 ) 2 + 2 AgCl
3
10.51. What percentage of 50.0 g of KClO 3 must be decomposed thermally to produce 14.6 g O 2 ?
Ans. The statement of the problem implies that not all the KClO 3 is decomposed; it must be in excess for the
purpose of producing the 14.6 g O 2 . Hence we can base the solution on the quantity of O 2 . In Problem 10.29
we found that 37.3 g of KClO 3 decomposed. The percent KClO 3 decomposed is then
37.3g
× 100% = 74.6%
50.0g
10.52. Percent yield is defined as 100 times the amount of a product actually prepared during a reaction divided by the
amount theoretically possible to be prepared according to the balanced chemical equation. (Some reactions are slow,
and sometimes not enough time is allowed for their completion; some reactions are accompanied by side reactions
which consume a portion of the reactants; some reactions never get to completion.) If 5.00 g C 6 H 11 Br is prepared
by treating 5.00 g C 6 H 12 with excess Br 2 , what is the percent yield? The equation is
C 6 H 12 + Br 2 −→ C 6 H 11 Br + HBr